Answer
$$f'(t)=\frac{4-t^2}{2\sqrt{\frac{t}{t^2+4}}(t^2+4)^2}$$
Work Step by Step
$$f'(t)=\left(\sqrt{\frac{t}{t^2+4}}\right)'=\frac{1}{2\sqrt{\frac{t}{t^2+4}}}\left(\frac{t}{t^2+4}\right)'=\frac{1}{2\sqrt{\frac{t}{t^2+4}}}\frac{t'(t^2+4)-t(t^2+4)'}{(t^2+4)^2}=\frac{1}{2\sqrt{\frac{t}{t^2+4}}}\frac{1\cdot(t^2+4)-t\cdot2t}{(t^2+4)^2}=\frac{4-t^2}{2\sqrt{\frac{t}{t^2+4}}(t^2+4)^2}$$
